## Precalculus (10th Edition)

$27$
$x^3-27=x^3+3x^2-3x^2+9x-9x-27=(x^3+3x^2+9x)-3x^2-9x-27=(x(x^2+3x+9))-3(x^2+3x+9)=(x^2+3x+9)(x-3)$ General formula for average rate of change from $x$ to $y$: $\frac{f(y)-f(x)}{y-x}.$ We do what we have to according to the exercise ($c=3$, $f(x)=x^3$): $$\lim_{x\to 3}\frac{f(x)-f(3)}{x-3}=\lim_{x\to 3}\frac{x^3-27}{x-3}=\lim_{x\to 3}\frac{(x-3)(x^2+3x+9)}{x-3}=\lim_{x\to 3}(x^2+3x+9)=3^2+3\cdot3+9=9+9+9=27.$$ The last part is true because $x^2+3x+9$ is continuous at $3$, so we can get its value at $x=3$ by plugging $x=3$ in.